# Eulers improved Formula

Announcements
#1
This question seems to be missing something for part (b). Either that or I do not know how to use an aspect of Euler's improved formula.

The normal form is to use part(a) in the question to answer (b). However I seem to need a when I have only and . In other words I have y(3) but I need a y(2.9) to use the improved formula.

I guessed maybe there was a typo and f(3.2) was needed but there is an answer in the book of a) 2.6238 and b) 2.6494. The substitution yields 2.6238 OK for part a but my f(3.2) is 3. something.

The Q is from p309 Q8 in the Cambridge A-Level Maths for AQA Student book 2 (Year 2)

The question is y(3)=2

a) Use the Euler formula with h=0.1, to obtain an approximation to y(3.1) to 4 dp.

(b) Use the improved Euler Formula with h=0.1, to obtain an approximation to y(3.1), to 4dp.
0
2 years ago
#2
(Original post by nerak99)
This question seems to be missing something for part (b). Either that or I do not know how to use an aspect of Euler's improved formula.

The normal form is to use part(a) in the question to answer (b). However I seem to need a when I have only and . In other words I have y(3) but I need a y(2.9) to use the improved formula.

I guessed maybe there was a typo and f(3.2) was needed but there is an answer in the book of a) 2.6238 and b) 2.6494. The substitution yields 2.6238 OK for part a but my f(3.2) is 3. something.

The Q is from p309 Q8 in the Cambridge A-Level Maths for AQA Student book 2 (Year 2)

The question is y(3)=2

a) Use the Euler formula with h=0.1, to obtain an approximation to y(3.1) to 4 dp.

(b) Use the improved Euler Formula with h=0.1, to obtain an approximation to y(3.1), to 4dp.
You need two starting points for b). y1 and y2 enable you to calculate y3 and hence march onwards.
You're using a central difference formula to approximate the derivative.
https://en.m.wikipedia.org/wiki/Nume...ifferentiation
Last edited by mqb2766; 2 years ago
0
#3
(Original post by mqb2766)
You need two starting points for b). y1 and y2 enable you to calculate y3 and hence march onwards.
You're using a central difference formula to approximate the derivative.
https://en.m.wikipedia.org/wiki/Nume...ifferentiation
OK what aqua refer to as Eulers improved method, is a central difference formula. I just used the name aqa used.
If we refer to the question I asked, in simple terms, “Is there an error in the question?” Do you think the answer is yes?

In other words, to use Eulers improved formula (aka ‘a central difference formula’) I would need more information than I have been given in the question.

That still leaves where the answer comes from?
Last edited by nerak99; 2 years ago
0
2 years ago
#4
(Original post by nerak99)
OK what aqua refer to as Eulers improved method, is a central difference formula. I just used the name aqa used.
If we refer to the question I asked, in simple terms, “Is there an error in the question?” Do you think the answer is yes?

In other words, to use Eulers improved formula (aka ‘a central difference formula’) I would need more information than I have been given in the question.

That still leaves where the answer comes from?
Can you upload an image of the full question?
0
#5
As requested.
0
#6
Sorry it is upside down! Hence my preference for latex
0
2 years ago
#7
(Original post by nerak99)
As requested.
Full question appears to be
https://pmt.physicsandmathstutor.com... - FP3 AQA.PDF
So yes, dodgy version of the question. Rather than doing a "vanilla" central difference as in the OP, they've (2010 exam) done a "predictor corrector" version where you first predict y(3.1), then use that to estimate the derivative at that point, then average the two derivatives to estimate the derivative in the center. Then use that to correct the original prediction of y(3.1).

Last edited by mqb2766; 2 years ago
0
#8
(Original post by mqb2766)
Full question appears to be
https://pmt.physicsandmathstutor.com... - FP3 AQA.PDF
So yes, dodgy version of the question. Rather than doing a "vanilla" central difference as in the OP, they've (2010 exam) done a "predictor corrector" version where you first predict y(3.1), then use that to estimate the derivative at that point, then average the two derivatives to estimate the derivative in the center.

Thanks very much for that.
1
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Were exams easier or harder than you expected?

Easier (21)
27.27%
As I expected (23)
29.87%
Harder (29)
37.66%
Something else (tell us in the thread) (4)
5.19%